\subsection*{2012-06-29}

  What is hydrodynamics? A dynamical theory that describes a liquid! How is it different from, say, free gas theory? Interactions! 

  A round-about way of characterizing the difference between a liquid and a gas is to say that the liquid obeys the hydrodynamical equation. 
  
  Setup. 
  \begin{enumerate}
    \item The liquid's density is approximately constant, $\rho_0 = \rho(x, t)$. 
    \item The liquid dynamics is completely described by the dynamics of the velocity field $\vec v(x, t)$ which is interpreted as velicities of infinitesimal volume elements. 
      Volume elements can stretch and shear, but they do not diffuse (change volume), in contrast to the case of gas. 
    \item The most general form of dynamical equation for the velocity is: 
    \begin{equation}
      \partial_t \vec v = F\left(\vec v,\, \vec \nabla \cdot \vec v,\, \vec \nabla^2 \vec v,\, (\vec v \cdot \vec \nabla)\, \vec v,\, \vec \nabla \times \vec v,\, \dots \right)
    \end{equation}
    We assume we're in the limit $\nabla v < 1$, expand the above equation in \emph{powers of $\nabla$}, and ignore terms of order $\nabla^2$ and higher. 
    This is called the hydrodynamics expansion.
  \end{enumerate}
  The fundamental assumption is that the system is completely described by the single velocity field $\vec v$. 
  This assumption may be found to be false, and in principle we would have to incorporate more dynamical quantities into our model (like a tensor instead of a vector, say). 

  Let's look at the resulting form of the hydrodynamics equation:
  \begin{equation}
    \partial_t \vec v = \alpha\; \vec v \left( \vec \nabla \cdot \vec v\right) + \beta \left( \vec v \cdot \vec \nabla \right) \vec v + \gamma \left( \vec \nabla \times \vec v \right) 
     + \zeta\; \vec \nabla^2 \vec v + \xi \left( \vec \nabla \cdot \vec v \right) ^2 \vec v
  \end{equation}  
  The equation just lists all possible ways $\nabla$ could appear at a certain power in the series expansion. 
  The various overall factors of $\vec v$ appear to make each quantity a vector. 
  Notice in principle we could have some non-trivial overall $f(\vec v)$ in front of each term. 
  For now ignore. 

  The coefficients $\alpha, \beta, \gamma, \zeta, \xi$ are called the transport coefficients. 
  We can attach physical meaning to them. 

  The first two terms, $\vec v \left( \vec \nabla \cdot \vec v\right )$ and $\left(\vec v \cdot \vec \nabla \right) \vec v$, both describe the divergence of $v$. 
  The first one -- total divergence, the second -- divergence in the direction of flow lines. 
  We expect both $\alpha$ and $\beta$ to be positive, to properly encode $\partial_t \vec v \propto \mathrm{div}\; \vec v$ (draw a picture to convince yourself).
  Supposedly, imposing some sort of symmetry (gallilean) constrains the relation between the two $\alpha$ and $\beta$. 

  The $\zeta$ coefficient is in fact the \emph{shear viscosity}. 
  From dimensional analysis, $\zeta \sim \tfrac{L^2}{T} \sim \tfrac{p \cdot x}{m}$. 
  Oh, was that an $m$? Uhm...
  The $\zeta$ term can be seen to be responsible for dissipation of energy in the system: imagine gradients vanish, then $\partial_t \vec v = \zeta\; \vec \nabla^2 \vec v$. 
  If $v_x \sim \sin kx$, then $\partial_t v_x = -k^2 \zeta v_x$ -- exponential decay $\rightarrow$ dissipation.

  The precise values of the coefficients differ from system to system. 
  Think water \& honey. 
  The general form of the equation is universal. 

  Notice the differential equation for $v$ is first order in $\partial_t$.
  This is completely different from classical mechanics, where the equations that follow from a classical Lagrangian are always second order and feature \emph{2 integration constants}. 
  Hydrodynamics is a phenomenological model that is not derived from first principles!
  
  What are the conserved quantities? Definitely not $v$.
  Energy? Momentum? Well, what \emph{is} energy in a model that doesn't follow from a lagrangian? 
  We need to figure out relations between $\epsilon$ (energy), $p$ (pressure), $\rho$ (density) as functions of $v$ (the only dynamical quantity) ourselves. 
  Again, the precise relation is model-dependent. 
  Historically (Navier-Stokes equation), people would figure out the conserved quantity from physical considerations: $Q = \epsilon u^2 + p \left( 1 - u^2 \right)$, plug it into the hydro equation and constrain the particular form.
  We do a similar thing, except that we're working in a relativistic framework.
  
  We postulate the existance of a conserved, as well as familiar, energy-momenty tensor $T^{\mu \nu}(u)$ ($u$ -- the relativistic velocity) that we express in terms of (yet unknown) $\epsilon$, $p$.
  There're $\tfrac{d\left(d+1\right)}{2}$ coefficients in a symmetric rank 2 tensor, but $u$ is only $d-1$ independent quantities (since we additionally require $u^2 = 1$).
  It will turn out we can add extra terms $\Pi^{\mu\nu}$ to $T$ that are independent of $u$, but Allan wasn't exactly clear on this -- read in literature. 
  So our $T$ will look like $T^{\mu \nu} = \epsilon u^\mu u^\nu + p \Delta^{\mu \nu}(u) + \Pi^{\mu \nu}$. 
  The $\Delta$ term is a term with tensor structure determined by symmetries (boh..?). 













\subsubsection*{Homework}

\begin{enumerate}
  \item Show that the $\zeta$ coefficient in the hydrodynamics expansion is in fact the shear viscosity. 
  \item Read read read. Check out all the references, look up unfamiliar terms in wiki and beyond, find more references. Landau has a good introduction to hydro. 
  \item In particular, read in Mark V. R. about 2+1 hydrodynamics (that will be dual to 3+1 AdS, but forget that part for now).
    Learn how to obtain equations for $\partial_t \vec v$ and $\partial_t \Pi^{\mu\nu}$ from the prescription outlined by Allan. 
\end{enumerate}


\subsubsection*{Talk about next time (tuesday)}
\begin{enumerate}
  \item The three amigos present everything they know on hydro. 
  \item Still want to hear about the other possible projects \& follow-up comuptation to our summer work! 
\end{enumerate}
